Understanding proof for $\sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$

Question

(English is not my native language, so I apologize if I fail to use the right technical terms)

I am stuck in proving the following. I'll explain how far I got and maybe someone can help me out by explaining the last part to me.

Show that:

$$\sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$$

So far so good. This is how far I got:

If $z$ is an $n$th root of unity, $z^n = 1$, thus $$0 = z^n - 1 = z^n - 1^n = (z - 1)\cdot \sum_{k=0}^{n-1}{z^k}$$
If $z \neq 1$ this means that $\sum_{k=0}^{n-1}{z^k} = 0$
Let $z$ be the root: $e^{\frac{2\pi \mathrm{i}}{n}}$, thus $$\sum_{k=0}^{n-1}{{e^{\frac{2\pi k \mathrm{i}}{n}}}} = 0$$
With Euler's formula, this is $$1 + \sum_{k=1}^{n-1}{\left(\cos{\frac{2\pi k}{n}} + \mathrm{i} \sin{\frac{2\pi k}{n}}\right)} = 0$$
Let's rearrange that a bit: $$1 + \sum_{k=1}^{n-1}{\cos{\frac{2\pi k}{n}}} + \mathrm{i} \sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$$

Okay, so this is where I got stuck. I thought this would somehow lead to a solution, but I couldn't figure it out. After a while, and trying different approaches, I looked at the sample solution: They do exactly what I am doing until step 3 (not as detailed) and then simply conclude with:

The imaginary part provides the desired equation.

Somehow I am unable to figure out why. All I can say from the equation above is that the cosine-sum needs to be $-1$ in order for the sine-sum to be 0. Am I missing something obvious here?

Thanks a lot for any help in explaining that!

If $a + ib = 0$ with real $a$ and $b$, then $a = 0$ and $b = 0$. Here, $b$ is your sum of sines. — Daniel Fischer, Aug 23 '14 at 20:52
Duh. Of course, how did I not see that? Thx. Unfortunately I cannot upvote comments. — johnnycrab, Aug 23 '14 at 20:57
@johnnycrab, See http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Aug 24 '14 at 05:11

score 2 · Answer 1 · answered Aug 23 '14 at 21:01

Non-complex answer: let $S=\sum_{k=1}^{n-1}\sin\frac{2\pi k}{n}$. If $n=1$, the problem is trivial so assume $n>1$. Multiply $S$ by (the nonzero factor) $\sin\frac{\pi}{n}$ and use the formula $$ \sin A\sin B=\frac{1}{2}(\cos(A-B)-\cos(A+B)). $$ You will get a telescoping sum that simplifies to give you: $$ \sin(\pi/n)S=\frac{1}{2}(\cos(\pi/n)-\cos((2n-1)\pi/n))=\frac{1}{2}(\cos(\pi/n)-\cos(2\pi-\pi/n))=0 $$ because $\cos(2\pi-\pi/n)=\cos(-\pi/n)=\cos(\pi/n)$.

score 0 · Accepted Answer · answered Aug 23 '14 at 20:53

0

Take the imaginary part of each side of the equation. On the left side you have your sum, on the right side you have zero.

answered Aug 23 '14 at 20:53

user157227

2,104

Understanding proof for $\sum_{k=1}^{n-1}{\sin{\frac{2\pi k}{n}}} = 0$

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